Question: The angles of elevation of the top of a tower from two points which are at distances of 10 m and 5 m from the base of the tower and in the same straight line with it are complementary. The height of the tower is
Options:
A) $ 5m $
B) $ 15m $
C) $ \sqrt{50}m $
D) $ \sqrt{75}m $
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Answer:
Correct Answer: C
Solution:
- Given that, angles are complementary.
Let h be the height of the tower.
Now, in $ \Delta PBC, $
$ \tan \theta =\frac{h}{5} $
(i)
and in $ \Delta PAC, $
$ \tan (90{}^\circ -\theta )=\frac{h}{10} $
$ \Rightarrow $ $ \cot \theta =\frac{h}{10} $
(ii)
On multiplying Eqs. (i) and (ii), we get
$ \tan \theta \cdot \cot \theta =\frac{h}{5}\times \frac{h}{10} $
$ \Rightarrow $ $ \frac{h^{2}}{50}=1 $
$ \Rightarrow $ $ h=\sqrt{50}m $
which is the required height of the tower.